函数中的评估范围

Question

我有以下脚本从字符串列表创建字典：

def dummy(a, b):
    c = a+1
    lst =["a","b","c"]
    return dict((k, eval(k)) for k in lst)

if __name__ == "__main__":
     dummy(0.01, "2015-01-29")

当我在终端中运行时，出现以下错误：

NameError: name 'a' is not defined

当eval起作用时，似乎在函数dummy的范围内找不到变量a...但我不明白为什么...

Answer 1

先列一个清单：

def dummy(a, b):
    c = a + 1
    lst =["a","b","c"]
    return dict([(k, eval(k)) for k in lst])

if __name__ == "__main__":
     print(dummy(0.01, "2015-01-29"))

您的版本是生成器。有关详细信息，请参阅此内容：

scope of eval function in python

引用相关部分：

The scope of names defined in a class block is limited to the class block; it does not extend to the code blocks of methods – this includes generator expressions since they are implemented using a function scope.

函数中的评估范围

Eval scope in a function

eval

python-2.7